In algebra a parabola — first of all the schedule of a square trinomial. However there is also a geometrical definition of a parabola as sets of all points which distance from some this point (parabola focus) is equal to distance to this straight line (head mistress of a parabola). If the parabola is set by the equation, then it is necessary to be able to calculate coordinates of its focus.
Instruction
1. Going from the return, suppose, that the parabola is set geometrically, that is its focus and the head mistress are known. For simplicity of calculations we will install the system of coordinates so that the head mistress was parallel to ordinate axis, focus lay on abscissa axis, and ordinate axis passed precisely in the middle between focus and the head mistress. Then the top of a parabola will coincide with the beginning of coordinates. In other words, if distance between focus and the head mistress to designate p, then coordinates of focus will be equal (p/2, 0), and the head mistress's equation — x = - p/2.
2. Distance from any point (x, y) to a point of focus it will be equal, on a distance formula between points, √ (x - p/2) ^2 + to y^2). The distance from the same point to the head mistress, respectively, will equal x + p/2.
3. Equating each other these two distances, you receive the equation: √ (x - p/2) ^2 + y^2) = x + p/2. Squaring both members of equation and removing the brackets, you receive: x^2 - px + (p^2)/4 + y^2 = x^2 + px + (p^2)/4. Having simplified expression, you will come to the final wording of the equation of a parabola: y^2 = 2px.
4. From this it is visible that if the equation of a parabola can be given to a type of y^2 = kx, then coordinates of its focus will be equal (k/4, 0). Having traded variables places, you will come to the algebraic equation of a parabola of y = (1/k) *x^2. Coordinates of focus of this parabola are equal (0, k/4).
5. The parabola serving as the schedule of a square trinomial is usually set by y equation = Ax^2 + Bx + C where A, B, and C — constants. The axis of such parabola is parallel to ordinate axis. The derivative of the square function set by Ax^2 trinomial + Bx + C is equal 2Ax + to B. She addresses in zero at x = - B/2A. Thus, coordinates of top of a parabola are equal (-B/2A, - B^2 / (4A) + C).
6. Such parabola is completely equivalent to the parabola set by y equation = Ax^2 shifted by parallel translation on - B/2A on abscissa axis and on-B^2 / (4A) + to C in ordinate axes. It is easy to check it replacement of coordinates. Therefore, if the top of the parabola set by square function is in a point (x, y), focus of this parabola is in a point (x, y + 1 / (4A).
7. Substituting the values of coordinates of top of a parabola calculated on the previous step to this formula and simplifying expressions, you finally receive: x = - B/2A, y = - (B^2 - 1) / 4A + C.